A radioactive mass emits particles according to a Poisson process at a mean rate of 3.5 per second. Let T be the waiting time, in seconds, between emissions.
(a) What is the mean waiting time?
(b) What is the median waiting time?
(c) Find P(T >2).
(d) Find P(T< 0.1).
(e) Find P(0.3 < T < 1.5).

(a) What is the mean waiting time?  = 0.875

(b) What is the median waiting time? =  0.594

(c) Find P(T >2).  = 0.102

(d) Find P(T< 0.1).  = 0.108

(e) Find P(0.3 < T < 1.5). = 0.53

Step-by-step explanation:

a)

Let the particles are emitted at a mean rate of 3.5 per second, So, the mean waiting time is 0.875

b)

From the formula, the mean of the exponential distribution is

μ = 1/λ = 0.875

    λ = 1.143

Let m denote the median

Then P(T≤m) = 0.875

Using the formula for the cumulative distribution function of the exponential distribution

P(T≤m) = 0.875

1 - = 0.875

    = 0.125

   - 3.5m = Ln (0.125)

         m = -2.079 / -3.5

         m = 0.594

Solving for m, we obtain that the median is 0.594

c)

for P(T >2)

P(T >2) = 1 - P(T≤2)

            = 1 - (1 - e^(-λ2))

            = e^(-λ2)

            = e^(-1.143x2)

            = e^(-2.286)

            = 0.102

d)

P(T<0.1) = 1 - e^(-λ*0.1)

             = 1 - e^(-1.143*0.1)

             = 1 - 0.892

             = 0.108

e)

P(0.3 < T < 1.5)

P(T< 1.5) = 1 - e^(-λ(1.5))

              = 1 - e^(-1.143(1.5))

              = 1 - 0.18

              = 0.82

P(T < 0.3) = 1 - e^(-λ(0.3))

                = 1 - e^(-1.143(0.3))

                = 1 - 0.71

                = 0.29

Therefore, P(0.3 < T < 1.5) = P(T< 1.5) - P(T < 0.3)

                                           = 0.82 - 0.29

                                           = 0.53

this is false ^.^ lol

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step-by-step explanation: